Cho biểu thức M=\(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
ĐK: x≥0
Tìm GTLN của
B= 1/M - \(\dfrac{\sqrt{x}}{27}\)
A= 1/M - \(\dfrac{\sqrt{x}+5}{12}\)
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Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)
Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi x=0 (tm)
Vậy \(A_{max}=\dfrac{1}{2}\)
Bài 2:
Đk: \(x\ge3;y\ge5;z\ge4\)
Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)
Áp dụng AM-GM có:
\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)
\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)
\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)
Cộng vế với vế \(\Rightarrow VT\ge20\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)
Vậy...
I miss you Được em, hoặc trực tiếp nhóm thành HĐT, một vế là tổng các bình phương, vế còn lại bằng 0
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
\(M=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\div\dfrac{\sqrt{x}-1}{2}\)
(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))
\(=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right]\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(x+2\right)+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\)
\(M=\dfrac{1}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le1\)
Dấu "=" xảy ra khi x = 0
Cảm ơn nhé! Nhưng tớ làm ra câu a,b rồi :( cậu biết làm c,d không?
Câu 1:
Tìm max:
Áp dụng BĐT Bunhiacopxky ta có:
\(y^2=(3\sqrt{x-1}+4\sqrt{5-x})^2\leq (3^2+4^2)(x-1+5-x)\)
\(\Rightarrow y^2\leq 100\Rightarrow y\leq 10\)
Vậy \(y_{\max}=10\)
Dấu đẳng thức xảy ra khi \(\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\Leftrightarrow x=\frac{61}{25}\)
Tìm min:
Ta có bổ đề sau: Với $a,b\geq 0$ thì \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
Chứng minh:
\(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
\(\Leftrightarrow (\sqrt{a}+\sqrt{b})^2\geq a+b\)
\(\Leftrightarrow \sqrt{ab}\geq 0\) (luôn đúng).
Dấu "=" xảy ra khi $ab=0$
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Áp dụng bổ đề trên vào bài toán ta có:
\(\sqrt{x-1}+\sqrt{5-x}\geq \sqrt{(x-1)+(5-x)}=2\)
\(\sqrt{5-x}\geq 0\)
\(\Rightarrow y=3(\sqrt{x-1}+\sqrt{5-x})+\sqrt{5-x}\geq 3.2+0=6\)
Vậy $y_{\min}=6$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-1)(5-x)=0\\ 5-x=0\end{matrix}\right.\Leftrightarrow x=5\)
Bài 2:
\(A=\sqrt{(x-1994)^2}+\sqrt{(x+1995)^2}=|x-1994|+|x+1995|\)
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(A=|x-1994|+|x+1995|=|1994-x|+|x+1995|\geq |1994-x+x+1995|=3989\)
Vậy \(A_{\min}=3989\)
Đẳng thức xảy ra khi \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)
\(\dfrac{1}{M}=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}-\dfrac{\sqrt{x}}{27}=\dfrac{27\sqrt{x}+54-x-5\sqrt{x}}{27\left(\sqrt{x}+5\right)}\)\(=\dfrac{-x+22\sqrt{x}+54}{27\left(\sqrt{x}+5\right)}\)
\(\Rightarrow\sqrt{x}.27B+135B=-x+22\sqrt{x}+54\)
\(\Leftrightarrow x+\sqrt{x}\left(27B-22\right)+135B-54=0\) (1)
Coi PT (1) là phương trình bậc 2 ẩn \(\sqrt{x}\)
PT (1) có nghiệm không âm \(\Leftrightarrow\left\{{}\begin{matrix}\Delta=729B^2-1728B+700\ge0\\S=22-27B\ge0\\P=135B-54\ge0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2}{5}\le B\le\dfrac{14}{27}\)
Suy ra \(max_B=\dfrac{14}{27}\Leftrightarrow x=16\)
A làm tương tự
Không làm được alo nha, giờ hành chính đến 0h30